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$X$ is a Hausdorff space and $\sim$ is an equivalence relation.

If the quotient map is open, then $X/ \sim$ is a Hausdorff space if and only if $\sim$ is a closed subset of the product space $X \times X$.

Necessity is obvious, but I don't know how to prove the other side. That is, $\sim$ is a closed subset of the product space $X \times X$ $\Rightarrow$ $X/ \sim$ is a Hausdorff space. Any advices and comments will be appreciated.

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I wonder if we can remove the condition that the quotient map is open. In that case, necessity is also obvious, is the sufficiency also true? Or is there any counterexamples? – user22111 Jan 2 '12 at 5:51
@Jingren: You should post this as a separate question. Anyway: Yes, the condition that the quotient map be open is necessary. Consider $X/A$ where $X$ is a non-regular Hausdorff space and $x$ is a point that cannot be separated from the closed set $A$. In the quotient $X/A$ the image of the point $x$ can't be separated from the point corresponding to $A$ while the equivalence relation is obviously closed. – t.b. Jan 2 '12 at 6:07
@yaoxiao Very nice post. – user38268 Apr 3 '12 at 13:54

Since the map $\pi:X\to X/\sim$ is open, it's clear that the map $g:X^2\to (X/\sim)^2$ given by $g(x,y)=(\pi(x),\pi(y))$ is open. What we claim is that $g(X^2-\sim)=(X/\sim)^2-\Delta_{X/\sim}$. Indeed, if $x\nsim y$ then $\pi(x)\ne\pi(y)$ which tells us that $g\left(X^2-\sim\right)\subseteq (X/\sim)^2-\Delta_{X/\sim}$. That said, if $(\pi(x),\pi(y))\notin\Delta_{X/\sim}$ then $\pi(x)\ne \pi(y)$ so that $x\nsim y$ so that $(x,y)\in X^2-\sim$ and clearly $g(x,y)=(\pi(x),\pi(y))$. Thus, $g(X^2-\sim)=(X/\sim)^2-\Delta_{X/\sim}$ as claimed. But, since $X^2-\sim$ is open by assumption, and $g$ is an open map we have that $(X/\sim)^2-\Delta_{X/\sim}$ is open, and so $\Delta_{X/\sim}$ is closed. This gives us $T_2$ness.

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Let $R$ be the subset of $X \times X$ which gives the equivalence relation $\sim$, and let $f\colon X \to X/{\sim}$ be the quotient map. Let $x, y \in X$ be points not equivalent under the relation, i.e. $(x, y) \notin R$. Since $R$ is closed and $X \times X$ has the product topology, there exist open sets $U, V$ in $X$ such that $(x, y) \in U \times V$ and $U \times V$ does not meet $R$. Can you separate $f(x)$ and $f(y)$ using $U$ and $V$? Remember that $f$ is assumed to be an open map.

[This is a lot like the proof of the fact that Alex is using: that a space $X$ is Hausdorff if and only if the diagonal is closed in $X \times X$.]

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Start with a point $(x,y)$ with $x$ and $y$ not related. Then, as the relation is reflexive, it contains the diagonal. Now, as the relation is closed, its complement is open and there is a neighbourood of $(x,y)$ which does not intersect it. Next think about what a base for the product topology might look like...

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Let $\pi:X\to X/\!\!\!\sim\;$ denote the projection map associated with $\sim$. (That is, for any $x\in X$, $\pi(x)$ is the $\sim$-equivalence class that $x$ belongs to.) Let $\nsim\; \subseteq X \times X$ be shorthand for the complement of $\;\;\sim\;\;$ in $X \times X\;$, i.e. $\nsim\;\;=\;(X \times X\;) \;\;-\; \sim\;$.

Suppose that $\pi(x) \neq \pi(y)\;$. (Here I'm relying on the fact that, since $\pi$ is surjective, any element $\widetilde{z}\in X/\!\!\!\sim\;$ may be written in the form $\pi(z)$, for some $z \in X$.) We must show that there exist open sets $U_{\pi(x)}, U_{\pi(y)} \subseteq X/\!\!\!\sim\;$ such that ${\pi(x)} \in U_{\pi(x)}$, ${\pi(y)} \in U_{\pi(y)}$, and $U_{\pi(x)} \cap U_{\pi(y)} = \varnothing\;$.

By assumption, $\;\sim\; \subseteq X \times X$ is closed, so $\nsim\; \subseteq X \times X$ is open. Therefore there exist open neighborhoods $N_x$ and $N_y$ of $x$ and $y$, respectively, such that $(x,\;y)\in N_x \times N_y \subseteq$$\;\;\nsim\;$. (This is because the family of all pairwise products of open subsets of $X$ is a basis for the product topology on $X \times X$.)

For any $v, w \in X$,

$$ (v,\;w) \;\in \;\nsim \;\;\;\;\;\Leftrightarrow\;\;\;\;\; \pi(v) \neq \pi(w)\;\;. $$


$$ N_x \times N_y \subseteq \;\;\nsim\;\;\;\;\Leftrightarrow\;\;\;\; \forall (v, w) \in N_x \times N_y \;[\pi(v) \neq \pi(w)] \;\;\;\;\Leftrightarrow\;\;\;\; \pi[N_x] \cap \pi[N_y] = \varnothing $$

Furthermore, since $\pi$ is open (by assumption), the image sets $\pi[N_x], \pi[N_y] \subseteq X/\!\!\!\sim\;$ are open neighborhoods of ${\pi(x)}$ and ${\pi(y)}$, respectively. Therefore, $\pi[N_x]$ and $\pi[N_y]$ are the desired open neighborhoods $U_{\pi(x)} \ni {\pi(x)}, U_{\pi(y)} \ni {\pi(y)}$.

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